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Sedaia [141]
3 years ago
15

A cylinder with a mass M and radius r is floating in a pool of liquid. The cylinder is weighted on one end so that the axis of t

he cylinder stands vertically. The total length of the cylinder is L. The density of the fluid is rho.
a. When the cylinder is in equilibrium, a portion of it is submerged in the liquid. What is the vertical length h of this submerged portion? Your answer should use only the given symbols M, r, L, and rho as necessary, as well as the constant g.
b. The cylinder is now displaced a distance y from the equilibrium position, such that |y| < h, and released from rest. Let the +y-axis be upward, so that if the cylinder is pulled up, y is positive, but if it is pushed down, y is negative. Immediately after it is released, what is the y-component of the acceleration ay of the cylinder? Your answer should use only the given symbols M, r, L, y, and rho as necessary, as well as the constant g.
c. The result of part (b) implies the cylinder is now in simple harmonic motion. What is the period T of the cylinder's oscillations
Physics
1 answer:
Blizzard [7]3 years ago
8 0

Answer:

acylinder whit is yoiyes

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A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
Harman [31]

Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

3 0
3 years ago
If a 10kg block is at rest on a table and a 1200N force is applied in the eastward direction for 10 seconds, what is the acceler
gavmur [86]

Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

m = 10 kg

Problem development

We replace the known data in formula (1)

1200 = 10*a

a = 1200/10

a = 120 m/s²

6 0
3 years ago
Science help please!
OLga [1]

Answer:

104 N

Explanation:

m = 1300 kg

a = 0.08m/s^2

F = 1300*0.08

F = 104 N

Newtons is the unit of force.

3 0
2 years ago
How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?​
grandymaker [24]

Answer:

-120000 W

Explanation:

Power = change in energy / time

P = ΔE / t

P = (½ mv₂² − ½ mv₁²) / t

P = m (v₂² − v₁²) / (2t)

Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:

P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)

P = -120000 W

7 0
3 years ago
Catalytic ozone destruction occurs in the stratosphere by a two-step reaction:
Sladkaya [172]
Catalytic ozone destruction occurs in the stratosphere where the reactions involving bromine, chlorine, hydrogen, nitrogen and oxygen gases form compounds that destroy the ozone layer. The reactions uses a catalyst (speeds up the reaction) in a two step reaction. considering chlorine the reactions appears as follows;
step 1
 Cl + O3 = ClO + O2 
 step 2
ClO + O = Cl + O2 
Where by chlorine is released to destroy the ozone layer, this takes place many times even with the other elements (hydrogen, bromine, nitrogen) and the end result is a completely destroyed Ozone layer
7 0
3 years ago
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